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Country Boy

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  1. You shouldn't have the decimal point in the numerator of that last fraction. Both 2105263157894736842105263157895 and 100000000000000000000000000000000 are divisible by 5 since the first ends in "5" and the second ends in 0. The first is 5 times 2105263157894736842105263157895 and the second is 5 times 20000000000000000000000000000000. Again those are both multiples of 5. The first of the two original numbers is 25 times 421052631578947368421052631579 and the second is 25 times 4000000000000000000000000000000. Unfortunately, the first of those two numbers does not end in 5 or 0 so is not divisible by 5. Since it is odd it is not divisible by 2. At this time we can pretty much stop. Since 4000000000000000000000000000000= 4(10^30)= 2^2(2^30 5^30)= 2^(32)5^30. The reduction to lowest terms of that fraction is 421052631578947368421052631579/4000000000000000000000000000000.
  2. Country Boy


    I would take "line" rather than "curve" to mean "straight line". Also, do not "hijack" someone else's thread to ask an unrelated question. Start your own thread. (As Strange said, in the Physics forum, not Math.)
  3. Yes, you can say that 1. side AB is congruent to side BD. Reason: Given. 2. side AC is congruent to side CD. Reason: Given. 3. side BC is congruent to itself. Reason: The "reflexive law". Any geometric object is congruent to itself. Last: 4. triangle ABC is congruent to triangle DBC. Reason: SSS.
  4. In interpreting what results? Biologist might measure the Ph of water where a known species is living to determine in what range of Ph that species can live. Or, with that knowledge, they might use the Ph to determine what species can live there.
  5. Here is my take on it: If a and b are any numbers, then (a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3. And 3ab(a+ b)= 3a^2b+ 3ab^2. Subtracting, (a+ b)^3- 3ab(a+ b)= a^3+ b^3. Let x= a+ b, m= 3ab, and n= a^3+ b^3. That equation becomes the reduced cubic, x^3+ mx= n ("reduced" because there is no x^2 term). So given a and b, we can construct a cubic equation that has a+ b as a solution. What about the other way? Given m and n, can we determine a and b and so x? Of course we can! From 3ab= m, b= m/3a. Then a^3+ b^3= a^3+ (m/3a)^3= n. Multiplying by a^3, (a^3)^2+ (m/3)^3= na^3. That is a quadratic equation in a^3: (a^3)^2- n(a^3)+ (m/3)^3= 0. Using the quadratic formula, a^3= (n+/- sqrt(n^2- 4(m/3)^3)/2. We can write that in a little nicer form as a^3= (n/2)+/- sqrt((n/2)^2- (m/3)^3). From a^3+ b^3= n, we have b^3= n- a^3= (n/2)-/+ sqrt((n/2)^2- (m/3)^3). The various combinations of "+" or "-" give the three roots of the cubic equation.
  6. In symbolic logic, any statement that starts "if (a false statement) then …" is true no matter whether the conclusion is true or false. You have begun several threads in which you start "if" followed by a false statement!
  7. The distance from 3 to 13 is 10. 4/5 of that is 10(4/5)= 8 so 4/5 of the way from 3 to 13 is 3+ 8= 11. The (signed) distance from -5 to -15 is -10. 4/5 of that is -10(4/5) is -8 so 4/5 of the way from -5 to -15 is -5+ (-15- (-5))(4/5)= -4+ (-10)(4/5)= -4- 8= -12. The point 4/5 of the way between (3, -5) to (13, -15) is (11, -12).
  8. After you have chosen a specific "xy" Cartesian coordinate system, then the set of (x, y) points satisfying $x^2+ y^2= r^2$ gives a circle with center at the origin of that coordinate system and radius r in the units used for that coordinate system.
  9. Given an open line segment, say (a, b), there are two ways to "compactify" it. One, the "Stone-Cech compactification", is to add the "endpoints", a and b, to get [a, b]. Another, the "one point compactification" is to imagine bending the line segment into a circle and adding a single point at the join. Following those ideas, we can "compactify" the set of all real numbers by adding "+ infinity" and "- infinity" or by adding a single point, "infinity", so that the topology becomes that of a circle, Similarly here are two different ways to "compactify" the open disk, [tex]\{(x, y)| x^2+ y^2< r^2\}[/tex]. The "Stone-Cech compactification", by adding the points on the boundary to get [tex]\{(x, y)| x^2+ y^2= r^2\}[/tex], and the "one point compactification" where you bend the disk up into a sphere, adding a single point at the top. We can do the same thing with the set of all complex numbers, adding an infinite number of "points at infinity", one in every direction, or add a single "point at infinity". In the first case the topology is that of a closed disk and in the second the topology is that of a sphere.
  10. The acceleration due to gravity is -g (which you are taking to be -10) vertically. Taking y to be positive upward and x positive to the right, the equations of motion are x= v_x t and y= v_y t- (g/2)t^2. The initial speed is 10 m/s at 30 degrees so v_x= 10(cos(30))= 10(1/2)= 5 m/s and v_y= 10(sin(30))= 10(sqrt(3)/2)= 5 sqrt(3) m/s which is about 8.7 m/s. So we have x= 5t and y= 8.7t- 5t^2. From x= 5t, t= x/5 so y= 8.7(x/5)- 5(x^2/25)= 1.74x- 0.2x^2. That is, of course, a parabola opening downward. The slope of the tangent line, at a given x, is 1.74- 0.4x. After 0.5 seconds, x= 5(0.5)= 2.5 and y= 8.7(0.5)- 5(0.25)= 5.6. The slope of the tangent line is 1.74- 0.4(2.5)= 0.74 so the tangent line to the trajectory at that time is y= 0.74(x- 2.5)+ 5.6. The tangential acceleration is the projection of the vector (0, -10) on that line.
  11. "It must not be vertebrate or fish". Fish are vertebrates!
  12. Help with what? What are you trying to do? Do you have a question?
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